40th Putnam 1979

z_i are complex numbers. Show that |Re[ (z₁² + z₂² + ... + z_n²)^1/2 ]| ≤ |Re z₁| + |Re z₂| + ... + |Re z_n|.

Solution

Let z_k = x_k + i y_k, and let (z₁² + z₂² + ... + z_n²)^1/2 = a + ib.

Then, squaring, a² - b² = ∑ x_k² - ∑ y_k² (1), ab = ∑ x_ky_k (2). The Cauchy inequality gives that |∑ x_ky_k| ≤ ( ∑ x_k² )^1/2 ( ∑ y_k² )^1/2, so from (1) we have |ab| ≤ ( ∑ x_k² )^1/2 ( ∑ y_k² )^1/2 (3). If |a| > ( ∑ x_k² )^1/2, then from (3) |b| < ( ∑ y_k² )^1/2. But then a² - b² > ∑ x_k² - ∑ y_k², contradicting (1). So we must have that |a| ≤ ( ∑ x_k² )^1/2. The required result follows immediately since obviously ( ∑ |x_k| )² ≥ ∑ x_k².

40th Putnam 1979

© John Scholes
jscholes@kalva.demon.co.uk
4 Dec 1999